Suppose that $f$ satisfies the equation $$f(x+y)=f(x)+f(y)+x^2y+xy^2$$for all real numbers $x$ and $y$. Suppose further that$$~~~~~~~~~~~~~\lim_{x\to0}\frac{f(x)}x=1$$a) Find $f(0)$
b) Find $f'(0)$
c) Find $f'(x)$

Question

Suppose that $f$ satisfies the equation $$f(x+y)=f(x)+f(y)+x^2y+xy^2$$for all real numbers $x$ and $y$. Suppose further that$$~~~~~~~~~~~~~\lim_{x\to0}\frac{f(x)}x=1$$a) Find $f(0)$
b) Find $f'(0)$
c) Find  $f'(x)$

turingtest · Answer

I found $f(0)=0$ but now I am not seeing how to proceed. Just a hint would be nice.

turingtest · Answer

I further think I proved that the function is odd, though I think that doesn't really help me :/

zarkon · Answer

use the definition of a derivative

zarkon · Answer

the fact that it is odd does seem to help with the last part

turingtest · Answer

I was thinking of using the definition of the derivative somehow but couldn't figure out how to get it into the right form....

zarkon · Answer

$$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}$$

turingtest · Answer

limit f(x)-f(0) something...

turingtest · Answer

ah dang but... oh, okay I guess I was trying to get things algebraically manipulated there, but I see now that since x=f(x) at x=0 the limit $is$ the definition
got it :)

turingtest · Answer

okay so that is f'(0)=1, on to the last part
you say the oddness could help?

zarkon · Answer

yes

turingtest · Answer

can I use the derivative of an odd function is even?

zarkon · Answer

don't need it

zarkon · Answer

use the definition of a derivative again

turingtest · Answer

that's what I'm doing, let me toy for a sec...

zarkon · Answer

I found f(x)  ;)

turingtest · Answer

haha, dang
okay then go ahead and give me a little more help on finding f'(x)
they don't even ask for f(x), jeez

turingtest · Answer

...if you would be so kind

zarkon · Answer

if you know f'(x) and that f(0)=0 then finding f(x) is really easy  :)

zarkon · Answer

$$f'(x)=\lim_{y\to x}\frac{f(y)-f(x)}{y-x}$$

zarkon · Answer

$$f(y)-f(x)=f(y)+f(-x)$$

zarkon · Answer

does that help?

zarkon · Answer

ok

kinggeorge · Answer

If it helps, it's easier for me to see it when I write the derivative as $$f'(x)=\lim_{y\to0} \frac{f(x+y)-f(x)}{y}$$

turingtest · Answer

that would be $$\lim_{y\to0}{f(y)+x^2y+xy^2\over y}=1+\lim...$$ah... it's $f'(x)=x^2+1$, right ?
:)

zarkon · Answer

yes

turingtest · Answer

sweet, thanks y'all
I still want to see how your way was gonna go @Zarkon  , but I'll work on that myself for a while

zarkon · Answer

gotta go...fun problem...I'll post it later if you like

turingtest · Answer

always a pleasure :)